Question

The displacement of a particle of a string carrying a travelling wave is given by y= (3.0 cm) sin 6.28(0.50x - 50 t),

where x is in centimeter and t in second. Find

(i) the amplitude (p) 50

(ii) the wavelength (q) 3

(iii) the frequency(r) 2

(iv) the speed (s) 100

• A. (i) - (r); (ii) - (p); (iii) - (s); (iv) - (q)
• B. (i) - (q); (ii) - (r); (iii) - (p); (iv) - (s)
• C. (i) - (p); (ii) - (q); (iii) - (r); (iv) - (s)
• D. (i) - (s); (ii) - (r); (iii) - (q); (iv) - (p)

Answer 1

The correct option is B

(i) - (q); (ii) - (r); (iii) - (p); (iv) - (s)

The standard wave equation for a wave is $y=Asin(kx-\omega t)$

Given equation $y=3sin6.28(0.5x-50t)$

$\Rightarrow y=3sin(6.28\times 0.5x-6.28\times 50t)$

Comparing we get

amplitude A = 3 cm

Right here, we can see that B is the only possible option, but we may as well calculate the other values too.

Wave number $k=\frac{2\pi }{\lambda }=6.28\times 0.5$

$\Rightarrow \lambda =\frac{2\pi }{k}=2cm$

Angular frequency $\omega =6.28\times 50$

Frequency $f=\frac{1}{T}=\frac{\omega }{2\pi }=\frac{6.28\times 50}{2\times 3.14}$

$f=50Hz$

wave speed =$\frac{\lambda }{T}=2\times 50=100\frac{cm}{s}$